Question

Calculate the ke of ejected electron when uv radiation of frequency 1.6 Ã10-15 strike the surface of potassium metal thresh hold frequency of potassium is 1.5Ã10 14

Answer 1

Frequency of uv light, fy = 1.6×10^15 Hz

Threshold frequency of Potassium, f0 = 1.5×10^14 Hz

Since the frequency (and hence energy) of the incident light is more than the threshold frequency, electron will be emitted from its surface.

The amount of energy which will be used to remove the electron is called work function, given by (h×f0)
The rest amount of energy of the incident light will be used to provide kinetic energy to the emitted electron.

So KE of electron = energy of incident light - work function of potassium

Or KE = h(fy) - h(f0)

Or KE = h×( fy- f0)

Or KE = 6.63×10^(-34) × (1.6×10^15 - 1.5×10^14)

Or KE = 6.63×10^(-20) × 14.5

Or KE = 9.61 × 10^(-19) J


KE of emitted electron is 9.61 × 10^(-19) J

Answer 2

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Frequency of uv light, fy = 1.6×10^15 Hz

Threshold frequency of Potassium, f0 = 1.5×10^14 Hz

Since the frequency (and hence energy) of the incident light is more than the threshold frequency, electron will be emitted from its surface.

The amount of energy which will be used to remove the electron is called work function, given by (h×f0)

The rest amount of energy of the incident light will be used to provide kinetic energy to the emitted electron.

So KE of electron = energy of incident light - work function of potassium

Or KE = h(fy) - h(f0)

Or KE = h×( fy- f0)

Or KE = 6.63×10^(-34) × (1.6×10^15 - 1.5×10^14)

Or KE = 6.63×10^(-20) × 14.5

Or KE = 9.61 × 10^(-19) J

KE of emitted electron is 9.61 × 10^(-19) J